Question: Ajay is researching how the population of his hometown has changed over time. Specifically, he learns his hometown had a population of $20{,}000$ in $1990$, and that the population has since increased by about $8\%$ every $3$ years. Ajay predicts that his town can only support a population of $50{,}000$. Ajay is relieved to see that population has not exceeded $50{,}000$ $t$ years after $1990$. Write an inequality in terms of $t$ that models the situation.
Explanation: The strategy This problem involves a population increasing by $8\%$ every $3$ years. So, to find the population over time, we repeatedly multiply the initial population, $20{,}000$, by $1.08$. [Why?] Because of this, we know we can model the situation with an exponential expression of the form $ab^x$, where $a$ is $20{,}000$ and $b$ is $1.08$. We now only need to find $x$, which represents the number of times the population has increased by $8\%$. Finding the exponent Suppose $t$ years have passed. Since the population increases by $8\%$ every three years, the number of times the initial population is multiplied by $1.08$ is $\dfrac{t}{3}$. Writing an inequality We can now replace $x$ in the original model with $\dfrac{t}{3}$. Therefore, the expression 20,000 ⋅ ( 1.08 ) t 3 20{,}000\cdot (1.08)\^{\frac{t}{3}} models the population of Ajay's hometown $t$ years after $1990$. Since we know that this population did not exceed $50{,}000$, we can write the following inequality. 20,000 ⋅ ( 1.08 ) t 3 ≤ 50,000 20{,}000\cdot (1.08)\^{\frac{t}{3}}\leq50{,}000 The answer An inequality that models the situation is 20000 ⋅ ( 1.08 ) t 3 ≤ 50000 20000\cdot (1.08)\^{\frac{t}{3}}\leq50000.